A factory worker accidentally releases a crate of mass m that was being held at

Question

A factory worker accidentally releases a crate of mass m that was being held at rest at the top of a ramp with length d which is inclined at an angle to the horizontal. The coefficient of kinetic friction between the crate and the ramp, and between the crate and the horizontal factory floor, is k. (a) How fast is the crate moving as it reaches the bottom of the ramp? (b) How far will it subsequently slide across the floor? (Assume that the crate's kinetic energy does not change as it moves from the ramp onto the floor.) State your answers in terms of the given variables (use g where applicable).

Explanation / Answer

Kinetic energy at bottom k = Utop + Wfriction

Utop = m g d sin theta

Wfriction = -fk d

fk = muk N = muk m g d cos theta

k = m g d (sin theta - muk cos theta) = 1/2 m v2

v = sqrt [2 g d (sin theta - muk cos theta)]

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N = m g

1/2 m v2 = muk m g d'

d' = v2 / 2 muk g

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